|2x + 3| ≤ 5Closed endpoints
x ∈ [−4, 1]
Absolute Value Inequalities Calculator
Find the full solution set. Understand the endpoints. See every interval.
Build your inequality
Real coefficientsINEQUALITY FORM · |ax + b| ≤ c|2x + 3| ≤ 5
Zero, negative bounds & constant cases
The operator matters when c ≤ 0. If a = 0, evaluate the constant inequality |b| compared with c; do not divide by a. Finite decimal inputs are interpreted exactly. At most 60 significant digits per entry; exponents −100 to 100. Input fractions such as 1/3 are not supported; exact fractional answers are.
Inside uses AND. Outside uses OR.
For a positive bound, ≤ or < keeps the middle interval; ≥ or > keeps the two outer rays.
For a positive bound, ≤ or < keeps the middle interval; ≥ or > keeps the two outer rays.
SOLUTION INTERVAL · |2x + 3| ≤ 5x ∈ [−4, 1]
−4 ≤ x ≤ 1 · both endpoints are included.
Exact interval notation
Solution regionInside interval
EndpointsBoth included
Boundary values−4 and 1
Your solution on the number line
Equal-unit scaleSET-BUILDER NOTATION{x ∈ ℝ | −4 ≤ x ≤ 1}
The highlighted segment contains every solution, including the two filled boundary points.
Step-by-step solution
Inside · AND|2x + 3| ≤ 5
1Write the compound inequality−5 ≤ 2x + 3 ≤ 5
2Subtract 3 from every part−8 ≤ 2x ≤ 2
3Divide every part by positive 2−4 ≤ x ≤ 1
SOLUTION INTERVAL[−4, 1]
Dividing by a negative coefficient reverses the inequality signs. Final endpoints are presented in increasing order.
Check any value
Substitution testx = 0 IS IN THE SOLUTION SET|2(0) + 3| = 3 ≤ 5
Do the boundary values belong?
At x = −4|−5| = 5 ≤ 5Included
At x = 1|5| = 5 ≤ 5Included
Change ≤ to < and the endpoints drop out.
The same bounds become the open interval (−4, 1). Equality is no longer allowed.
The same bounds become the open interval (−4, 1). Equality is no longer allowed.
Four operators. Four different sets.
Same a = 2, b = 3 and c = 5 · compare the solution regions.
|2x + 3| < 5Open endpoints
x ∈ (−4, 1)
|2x + 3| ≥ 5Closed endpoints
x ∈ (−∞, −4] ∪ [1, ∞)
|2x + 3| > 5Open endpoints
x ∈ (−∞, −4) ∪ (1, ∞)
Filled point: includedOpen point: excluded∞ always takes a parenthesis.
Special cases are part of the solution
Check before dividing< or ≤ gives ∅.
> or ≥ gives ℝ.
< gives ∅; ≤ gives {−b/a}.
> excludes −b/a; ≥ gives ℝ.
Check |b| against c using the chosen operator. True means ℝ; false means ∅.
Formula & interpretation
For c > 0 and a ≠ 0:
center h = −b/a · radius r = c/|a|
center h = −b/a · radius r = c/|a|
Inside: h − r ≤ x ≤ h + r.
Outside: x ≤ h − r or x ≥ h + r.
Use strict inequalities and open endpoints for < or >.
An equality gives points. An inequality gives a set.
Here every x between −4 and 1 works, not just the boundary values.
Here every x between −4 and 1 works, not just the boundary values.
Frequently asked questions
When do I use brackets or parentheses?
Use a square bracket at a finite included endpoint and a parenthesis at an excluded endpoint. Infinity is never included.
Why does the outside case use a union?
The solution contains two separate rays: values below the lower bound or above the upper bound. The union symbol combines them.
What changes when a is negative?
Dividing an inequality by a negative number reverses its direction. The final solution is still written with ordered endpoints.
Does a negative bound always mean no solution?
No. Since absolute values are nonnegative, a negative bound makes < and ≤ impossible, but > and ≥ true for all real x.
Sources & Reference Guides
OpenStaxIntermediate Algebra · Absolute Value InequalitiesCompound inequalities, unions and special cases.OpenStaxAlgebra · Interval NotationOpen and closed endpoints on the real number line.
Independent educational references · No institutional endorsement or expert review claimed.
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Find the full solution set. Understand the endpoints. See every interval.